Discussion
A and B are playing mathematical puzzles. A asks B "the whole numbers, greater than one, which can divide all the nine three digit numbers 111,222,333,444,555,666, 777,888 and 999?"
B immediately gave the desired answer. It was:
Answer: B
Every one of the nine numbers is a multiple of 111:
\[ 111,\ 222 = 2 \times 111,\ \ldots,\ 999 = 9 \times 111 \]
So any common divisor of all nine must divide 111, and conversely every divisor of 111 divides them all.
\[ 111 = 3 \times 37 \]
The divisors greater than 1 are therefore 3, 37 and 111. 9 is excluded because \( 111 \div 9 \) is not whole.
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